In a first order reaction,87.5 % of the react | vedclass

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Question

(A) For a first order reaction,the amount of reactant remaining after $87.5 \%$ conversion is $100 \% - 87.5 \% = 12.5 \%$.Let the initial concentrati

Vedclass · Accepted Answer

$\frac{0.693}{5} \ min^{-1}$

Vedclass · Answer

(A) For a first order reaction,the amount of reactant remaining after $87.5 \%$ conversion is $100 \% - 87.5 \% = 12.5 \%$.Let the initial concentration $[A]_0 = 100$ and the concentration at time $t = 15 \ min$ be $[A]_t = 12.5$.The number of half-lives $n$ can be calculated as $12.5 = 100 	imes (1/2)^n$,which gives $(1/2)^n = 1/8$,so $n = 3$.Since $t = n 	imes t_{1/2}$,we have $15 = 3 	imes t_{1/2}$,which means $t_{1/2} = 5 \ min$.The rate constant $k$ is given by $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{5} \ min^{-1}$.

In a first order reaction,$87.5 \%$ of the reactant is converted into the product in $15 \ minutes$. The rate constant for the reaction is given by:

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